MCQMediumJEE Main 2025 · 7 April, Shift 2Conic Sections (Parabola, Ellipse, Hyperbola)

Mathematics Question from JEE Main 2025 · 7 April, Shift 2

Let e1e_1 and e2e_2 be the eccentricities of the ellipse x2b2+y225=1\frac{x^2}{b^2} + \frac{y^2}{25} = 1 and the hyperbola x216y2b2=1,\frac{x^2}{16} - \frac{y^2}{b^2} = 1, respectively. If b<5b < 5 and e1e2=1e_1 e_2 = 1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

  • A

    45\frac{4}{5}

  • B

    35\frac{3}{5}

  • C

    74\frac{\sqrt{7}}{4}

  • D

    32\frac{\sqrt{3}}{2}

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Conic Sections (Parabola, Ellipse, Hyperbola) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions